Each interior angle of a regular polygon is reduced by 4 degrees when the number of sides is decreased by 1 (maintaining its regular shape). How many sides does the polygon with more sides have?
I didn’t quite understand what it’s asking. With more sides than what?
Furthermore, in the solution, I think there are some missing reasoning steps (between “hence we have” and the equation) that is making me confused
You have a regular polygon with n sides. Its interior angle is x^{\circ}.
You have another regular polygon with n - 1 sides. Its interior angle is (x - 4)^{\circ}.
I see now. Let interior angle of polygon with n sides be x^{\circ} and let interior angle of polygon with n-1 sides be (x-4)^{\circ}. To solve for n, find the difference between the two polygons (using the interior angle formula):
\frac{180(n-2)}{n} - \frac{180((n-1)-2)}{(n-1)} = x - (x-4) \frac{180(n-2)}{n} - \frac{180((n-1)-2)}{(n-1)} = x -x+4) \frac{180(n-2)}{n} - \frac{180((n-1)-2)}{(n-1)} = 4
what am I missing here: (n-2)*180 = original sum = S. (n-2-1)180 = S - 4n. N=45. If we reduce each angle of n sided polygon by 4, aren’t we reducing the total sum by 4?
I see part of the problem. With 43 sides don’t make a polygon, but what am I missing? wouldn’t reducing 4 degrees for each angle, reduce sum of angles by 4n
aren’t number of angles. = number of sides of a polygon, so total sum of angles is 100, reduce all angles by 4 degrees, it would imply 100 - 4*number of sides?
Number of angles is a measure each angle if we know the sum. However, I am think if we are reducing each angle, but we are also loosing a side. Which means I can’t have sum of old polygon related to sum of new polygon by simply subtracting it by 4n. And the new regular polygon can never have the same number of sides as old with different angle. Am I right?
